Tough Cordinates - I

The center of the circle will be at the point of intersection of the perpendicular bisectors of any two sides of the triangle.

The side joining vertices $(2,3)$ and $(3,4)$ has slope $1$ and midpoint $(\frac{5}{2}, \frac{7}{2})$ . The perpendicular bisector of this side will then have slope $-1$ and pass through the midpoint, and thus has the equation

$y - \frac{7}{2} = -1*(x - \frac{5}{2}) \Longrightarrow y = -x + 6$ .

The side joining vertices $(3,4)$ and $(6,8)$ has slope $\frac{4}{3}$ and midpoint $(\frac{9}{2}, 6)$ . The perpendicular bisector of this side will then have slope $-\frac{3}{4}$ and pass through the midpoint, and thus has the equation

$y - 6 = (-\frac{3}{4})(x - \frac{9}{2}) \Longrightarrow y = (-\frac{3}{4})x + \frac{75}{8}$ .

These two perpendicular bisectors intersect when

$-x + 6 = (-\frac{3}{4})x + \frac{75}{8} \Longrightarrow (-\frac{1}{4})x = \frac{27}{8} \Longrightarrow x = -\frac{27}{2}$ ,

which in turn gives us $y = -(-\frac{27}{2}) + 6 = \frac{39}{2}$ .

Thus the center of the circle is the point $(-\frac{27}{2}, \frac{39}{2})$ .

Say the center was (x,y) the by applying distance formula between center and the first two points namely, (2,3) and (3,4) and upon solving we get an equation x+y=6. As we see that the options aren't that confusing I guess checking is best approach after this step.

I know this method is not full proof but still as I mentioned to get the correct answer I guess checking can be done here. However, I was also wondering if this was a level 4 question.

(midpoint of (3,4) and (6,8) is (4.5,6). Slope of the chord joining these two points is 4/3. So the perpendicular from the center to this chord would be Y= - 3/4(X)+3/4*(9/2)+6=-3/4X + 75/8.
Putting the X values given in this equation, only the third point gives correct Y. So that is the center.
This is only a short cut for this problem. Not a full general method.

since circle x^ 2 + y^ 2 + 2gx +2fy +c=0 circumscribes triangle..........the co.ordinates of triangle passes through the circle....so substitute all the 3 co ordinates of triangle in eq of circle formula......we will get 3 eq....after solving all 3 eq.....we will get value of g and f......then centre of circle is( -g,-f)..